Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  int flag2 = 0;
  if (n > m) {
    int temp = n;
    n = m;
    m = temp;
    flag2 = 1;
  }
  int a[n + 2][m + 2];
  if (flag2 == 1) {
    for (long long int j = (1); j <= (m); j++) {
      for (long long int i = (1); i <= (n); i++) {
        cin >> a[i][j];
      }
    }
    for (long long int i = (1); i <= (n); i++) {
      a[i][m + 1] = a[i][1];
      for (long long int j = (2); j <= (m); j++) {
        if (a[i][m + 1] > a[i][j]) a[i][m + 1] = a[i][j];
      }
    }
  } else {
    for (long long int i = (1); i <= (n); i++) {
      int minv = 501;
      for (long long int j = (1); j <= (m); j++) {
        cin >> a[i][j];
        if (a[i][j] < minv) minv = a[i][j];
      }
      a[i][m + 1] = minv;
    }
  }
  int ans = 0;
  vector<int> row;
  vector<pair<int, int> > col;
  for (long long int i = (1); i <= (n); i++) {
    if (a[i][m + 1] != 0) {
      row.push_back(i);
      ans += a[i][m + 1];
      for (long long int j = (1); j <= (m); j++) a[i][j] -= a[i][m + 1];
    }
  }
  for (long long int j = (1); j <= (m); j++) {
    int m = a[1][j];
    if (m != 0) {
      col.push_back(make_pair(j, m));
      ans += m;
      for (long long int i = (1); i <= (n); i++) a[i][j] = a[i][j] - m;
    }
  }
  int flag = 0;
  for (long long int i = (1); i <= (n); i++) {
    for (long long int j = (1); j <= (m); j++) {
      if (a[i][j] != 0) {
        flag = 1;
        break;
      }
    }
    if (flag == 1) break;
  }
  if (flag == 1)
    cout << "-1\n";
  else {
    cout << ans << "\n";
    int q = row.size() - 1;
    for (int i = 0; i <= q; i++) {
      while (a[row[i]][m + 1] > 0) {
        if (flag2 == 1)
          cout << "col " << row[i] << "\n";
        else
          cout << "row " << row[i] << "\n";
        (a[row[i]][m + 1]) -= 1;
      }
    }
    int y = col.size();
    for (long long int i = (0); i <= (y - 1); i++) {
      while (col[i].second > 0) {
        if (flag2 == 1)
          cout << "row " << col[i].first << "\n";
        else
          cout << "col " << col[i].first << "\n";
        col[i].second -= 1;
      }
    }
  }
  return 0;
}